Field Capacity Calculator: Acres per Hour, Hours, Days, Fuel and Cost

How it works

Effective field capacity (acres/hour) = width (ft) × speed (mph) × field efficiency ÷ 8.25 · metric: ha/h = width (m) × speed (km/h) × efficiency ÷ 10 · 8.25 = 43,560 ft² per acre ÷ 5,280 ft per mile · hours = area ÷ capacity · fuel = hours × fuel per hour · cost per acre = (fuel cost + hourly rate × hours) ÷ acres · capacity needed = area ÷ (calendar days × share of workable days × hours per day)

A machine covers a strip as wide as its working width at the speed it travels. Width in feet times speed in miles per hour, divided by 8.25, is the theoretical field capacity in acres per hour (8.25 is 43,560 square feet per acre divided by 5,280 feet per mile; in metric, metres times km/h divided by 10 gives hectares per hour). No machine works all the time, so the theoretical figure is multiplied by the field efficiency: the share of time spent actually working rather than turning on headlands, overlapping, filling seed, spray or fertilizer, unloading grain or wrapping bales. Typical efficiencies from Iowa State University Extension and Purdue are about 83% for tillage, 65% for planters up to 12 rows (falling to 55% at 48 rows), 68–70% for drills, 63–65% for pull-type sprayers and 58–60% for wide self-propelled ones, 75–82% for combines, 80% for mowers and rakes, 75% for rectangular balers, 65% for round balers and 68% for spreaders. The calculator picks the Iowa figure for your width. Long, square fields, auto-steer and good tender or grain-cart logistics push the figure up; small, irregular fields pull it down. To measure your own speed, time a pass over 88 feet: mph = 60 ÷ seconds. Capacity turns into hours and days for the job, fuel and cost per acre. With the optional window inputs, the calculator also works out the capacity you need: the area divided by the working hours in the days that are likely to be fit for field work. That is the question that decides whether the machine fits the farm, because missing the planting or harvest window costs yield.

Worked example

With the defaults: a 30 ft (12-row, 30-inch) planter at 5.5 mph and Iowa State's typical 65% field efficiency for that size, 640 acres, 10 hours a day, 8 gal/h of fuel at $3.60/gal and $120/h for machine and operator. Theoretical capacity = 30 × 5.5 ÷ 8.25 = 20.00 acres/hour; effective capacity = 20.00 × 65% = 13.0 acres/hour (5.26 ha/h), the same as Iowa State's table for a 12-row planter. 640 ÷ 13.0 = 49.2 hours, or 4.9 days at 10 hours a day. Fuel = 49.2 × 8 = 394 gal ($1,418), machine and labour = $5,908, so the cost is $11.45 per acre.

Sources: Iowa State University Extension AgDM A3-24: Estimating field capacity of farm machines (equations, speed from 88 ft) · Iowa State PM 696 Table 1: typical speeds, field efficiencies and capacities (FFA machinery skill reprint) · Purdue (Buckmaster): Selecting forage machinery, typical speeds and field efficiencies, required capacity

Frequently asked questions